Calculus Examples

$\int \frac{1 + x}{1 - x} d x$

Step 1

Reorder $1$ and $x$ .

$\int \frac{x + 1}{1 - x} d x$

Step 2

Reorder $1$ and $- x$ .

$\int \frac{x + 1}{- x + 1} d x$

Step 3

Divide $x + 1$ by $- x + 1$ .

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Step 3.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x$

$1$

Step 3.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $- x$ .

				-	$1$
-	$x$	+	$1$		$x$	+	$1$

Step 3.3

Multiply the new quotient term by the divisor.

				-	$1$
-	$x$	+	$1$		$x$	+	$1$
				+	$x$	-	$1$

Step 3.4

The expression needs to be subtracted from the dividend, so change all the signs in $x - 1$

				-	$1$
-	$x$	+	$1$		$x$	+	$1$
				-	$x$	+	$1$

Step 3.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				-	$1$
-	$x$	+	$1$		$x$	+	$1$
				-	$x$	+	$1$
						+	$2$

Step 3.6

The final answer is the quotient plus the remainder over the divisor.

$\int - 1 + \frac{2}{- x + 1} d x$

Step 4

Split the single integral into multiple integrals.

$\int - 1 d x + \int \frac{2}{- x + 1} d x$

Step 5

Apply the constant rule.

$- x + C + \int \frac{2}{- x + 1} d x$

Step 6

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$- x + C + 2 \int \frac{1}{- x + 1} d x$

Step 7

Let $u = - x + 1$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 7.1

Let $u = - x + 1$ . Find $\frac{d u}{d x}$ .

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Step 7.1.1

Rewrite.

$\frac{- 1}{1}$

Step 7.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 7.2

Rewrite the problem using $u$ and $d u$ .

$- x + C + 2 \int \frac{- 1}{u} d u$

Step 8

Move the negative in front of the fraction.

$- x + C + 2 \int - \frac{1}{u} d u$

Step 9

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$- x + C + 2 (- \int \frac{1}{u} d u)$

Step 10

Multiply $- 1$ by $2$ .

$- x + C - 2 \int \frac{1}{u} d u$

Step 11

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$- x + C - 2 (\ln (| u |) + C)$

Step 12

Simplify.

$- x - 2 \ln (| u |) + C$

Step 13

Replace all occurrences of $u$ with $- x + 1$ .

$- x - 2 \ln (| - x + 1 |) + C$